Suppose that $G$ is a group and $H$ is its subgroup such that for every subgroup $K$ of $G$, $HK$ is a subgroup. Can we deduce that $H$ is normal?
(I know that the normality of either $H$ or $K$ is sufficient for $HK$ to be normal, but not a necessity.)
I feel like(!) there is a counter-example, but I can't find one. Any help would be appreciated. :)
Edit: $HK = \{ hk: h \in H, k \in K \}$
On
From the giving condition, for any $\;x\in G\;$ we have that
$$\;\langle x\rangle H\;\;\text{is a subgroup}\;\;\iff \langle x\rangle H=H\langle x\rangle$$
and this means, that $\;xH=Hx^k\;$, for some $\;k\in\Bbb Z\;$ , and this means $\;H\lhd G\;$ according to this classical exercise
Subgroups with that property are called permutable or qusainormal. There are examples of permutatble subgroups that are not normal, although they are always subnormal. An example is a non-central subgroup of order $p$ in a nonabelian group of order $p^3$ and exponent $p^2$ for an odd prime $p$. That is, the subgroup $\langle b \rangle$ in $\langle a,b \mid a^{p^2}=b^p=1, [a,b]=a^p \rangle$.